Question

The sum of three numbers in A.P. is $-3$, and their product is $8$. Find the numbers.

Answer 1

Let the numbers be $(a-d),a,(a+d)$.

Then, Sum of numbers $-3⟹(a-d)+a+(a+d)=-3⟹3a=-3⟹a=-1$

Now,Product of numbers $=8$

$⟹(a-d)\left(a\right)(a+d)=8$
$⟹a(a^2-d^2)=8$
$⟹(-1)(1-d^2)=8$ $[\because a=-1]$
$⟹d^2=9⟹d=\pm 3$
If $d=3$, the numbers are $-4,-1,2$.

If $d=-3$, the numbers are $2,-1,-4$.

Thus, the numbers are $-4,-1,2$ or $2,-1,-4$.